I am reading this paper which on p.5 describes a condition for a linear operator A on a Hilbert space to be idempotent as $m(A \otimes A)m^*=A$, where m is the multiplication map defined by $m(a_w \otimes a_v)=\delta_{w,v}a_w$ and $m^*$ is its adjoint. It is claimed that this definition reduces to A being idempotent under pointwise multiplication, and the author provides the computation
$ m(A_1 \otimes A_2)m^*(e_v) = m(A_1(e_v) \otimes A_2(e_v)) = \sum_{u,w}A_{u,v}^{(1)}A_{w,v}^{(2)}m(e_u \otimes e_w) = \sum_u A_{u,v}^{(1)}A_{u,v}^{(2)}e_u$
I am confused about
(1) what is going on in the last two steps of this computation (where does the summation come from?),
(2) why idempotency is defined in this way rather than just saying $AA=A$ where $AA$ are point-wise multiplied. I assume it has something to do with subtleties concerning the basis of the Hilbert space but not sure what it is precisely.
The last 2 summation terms are coming from the following: $m(A(e_v) \otimes A(e_v)) = m(\sum_u A_{uv} e_u \otimes \sum_w A_{wv} e_w) = \sum_{u,w} m(A_{uv} e_v \otimes A_{wv} e_w) = \sum_{u,w} A_{uv} A_{wv} m( e_v \otimes e_w) = \sum_{u,w} A_{uv} A_{wv} \delta_{uw} e_u = \sum_{u} A_{uv} A_{uv} e_u = \text{for idempotency} = \sum_{u} A_{uv} e_u $
where $Ae_v = \sum_u A_{uv} e_u$.